Is it possible to compare two binary trees in less than O(n log n) time?












10














I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time. 



 public class TreeNode {

  int val;

  TreeNode left;

  TreeNode right;

  TreeNode(int x) { val = x; }

 }



 class Solution {

  public boolean isSameTree(TreeNode p, TreeNode q) {



    if  ( p == null && q==null)

        return true;



    if (p == null || q == null) 

        return false;



    if ( (p.val == q.val) && isSameTree(p.left, q.left) && 

      isSameTree(p.right, q.right))

        return true;

    else 

        return false;  

   }   

  }


My code takes O(n log n) time.



How to approach reducing the time required?






java algorithm time-complexity binary-tree







share|improve this question




















  • 1




    If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    yesterday






  • 4




    Don't write, never present uncommented code. Never code if (condition) return true; else return false;. Just // same tree if same root, left, and right return p == q || null != p && null != q && p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    – greybeard
    yesterday








  • 1




    What do you count as n? Your algorithm looks very much linear on the number of nodes.
    – Bergi
    yesterday
















10














I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time. 



 public class TreeNode {

  int val;

  TreeNode left;

  TreeNode right;

  TreeNode(int x) { val = x; }

 }



 class Solution {

  public boolean isSameTree(TreeNode p, TreeNode q) {



    if  ( p == null && q==null)

        return true;



    if (p == null || q == null) 

        return false;



    if ( (p.val == q.val) && isSameTree(p.left, q.left) && 

      isSameTree(p.right, q.right))

        return true;

    else 

        return false;  

   }   

  }


My code takes O(n log n) time.



How to approach reducing the time required?






java algorithm time-complexity binary-tree







share|improve this question




















  • 1




    If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    yesterday






  • 4




    Don't write, never present uncommented code. Never code if (condition) return true; else return false;. Just // same tree if same root, left, and right return p == q || null != p && null != q && p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    – greybeard
    yesterday








  • 1




    What do you count as n? Your algorithm looks very much linear on the number of nodes.
    – Bergi
    yesterday














10












10








10


1





I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time. 



 public class TreeNode {

  int val;

  TreeNode left;

  TreeNode right;

  TreeNode(int x) { val = x; }

 }



 class Solution {

  public boolean isSameTree(TreeNode p, TreeNode q) {



    if  ( p == null && q==null)

        return true;



    if (p == null || q == null) 

        return false;



    if ( (p.val == q.val) && isSameTree(p.left, q.left) && 

      isSameTree(p.right, q.right))

        return true;

    else 

        return false;  

   }   

  }


My code takes O(n log n) time.



How to approach reducing the time required?






java algorithm time-complexity binary-tree







share|improve this question















I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time. 



 public class TreeNode {

  int val;

  TreeNode left;

  TreeNode right;

  TreeNode(int x) { val = x; }

 }



 class Solution {

  public boolean isSameTree(TreeNode p, TreeNode q) {



    if  ( p == null && q==null)

        return true;



    if (p == null || q == null) 

        return false;



    if ( (p.val == q.val) && isSameTree(p.left, q.left) && 

      isSameTree(p.right, q.right))

        return true;

    else 

        return false;  

   }   

  }


My code takes O(n log n) time.



How to approach reducing the time required?






java algorithm time-complexity binary-tree




java algorithm time-complexity binary-tree






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









nullpointer

44k1095182




44k1095182










asked yesterday









LouiseLouise

542




542








  • 1




    If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    yesterday






  • 4




    Don't write, never present uncommented code. Never code if (condition) return true; else return false;. Just // same tree if same root, left, and right return p == q || null != p && null != q && p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    – greybeard
    yesterday








  • 1




    What do you count as n? Your algorithm looks very much linear on the number of nodes.
    – Bergi
    yesterday














  • 1




    If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    yesterday






  • 4




    Don't write, never present uncommented code. Never code if (condition) return true; else return false;. Just // same tree if same root, left, and right return p == q || null != p && null != q && p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    – greybeard
    yesterday








  • 1




    What do you count as n? Your algorithm looks very much linear on the number of nodes.
    – Bergi
    yesterday








1




1




If you happen to have a size variable at the base of the structure, compare that first.
– Boann
yesterday




If you happen to have a size variable at the base of the structure, compare that first.
– Boann
yesterday




4




4




Don't write, never present uncommented code. Never code if (condition) return true; else return false;. Just // same tree if same root, left, and right return p == q || null != p && null != q && p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
– greybeard
yesterday






Don't write, never present uncommented code. Never code if (condition) return true; else return false;. Just // same tree if same root, left, and right return p == q || null != p && null != q && p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
– greybeard
yesterday






1




1




What do you count as n? Your algorithm looks very much linear on the number of nodes.
– Bergi
yesterday




What do you count as n? Your algorithm looks very much linear on the number of nodes.
– Bergi
yesterday












1 Answer
1






active

oldest

votes


















16














The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.






share|improve this answer



















  • 9




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    yesterday










  • @StephenC big-O notation implies "worst case".
    – David M
    17 hours ago










  • It doesn't always. For example, we don't say that a HashMap insertion is O(N)! By convention, big O refers to the "average" or "typical" case if you don't qualify it. The problem is that for the OP's example, it is difficult to say what the average is. That is why it is a good idea to be explicit, rather than relying on some vague convention to interpret what we mean,
    – Stephen C
    13 hours ago













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









16














The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.






share|improve this answer



















  • 9




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    yesterday










  • @StephenC big-O notation implies "worst case".
    – David M
    17 hours ago










  • It doesn't always. For example, we don't say that a HashMap insertion is O(N)! By convention, big O refers to the "average" or "typical" case if you don't qualify it. The problem is that for the OP's example, it is difficult to say what the average is. That is why it is a good idea to be explicit, rather than relying on some vague convention to interpret what we mean,
    – Stephen C
    13 hours ago


















16














The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.






share|improve this answer



















  • 9




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    yesterday










  • @StephenC big-O notation implies "worst case".
    – David M
    17 hours ago










  • It doesn't always. For example, we don't say that a HashMap insertion is O(N)! By convention, big O refers to the "average" or "typical" case if you don't qualify it. The problem is that for the OP's example, it is difficult to say what the average is. That is why it is a good idea to be explicit, rather than relying on some vague convention to interpret what we mean,
    – Stephen C
    13 hours ago
















16












16








16






The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.






share|improve this answer














The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.







share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday









ruakh

124k13197251




124k13197251










answered yesterday









nullpointernullpointer

44k1095182




44k1095182








  • 9




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    yesterday










  • @StephenC big-O notation implies "worst case".
    – David M
    17 hours ago










  • It doesn't always. For example, we don't say that a HashMap insertion is O(N)! By convention, big O refers to the "average" or "typical" case if you don't qualify it. The problem is that for the OP's example, it is difficult to say what the average is. That is why it is a good idea to be explicit, rather than relying on some vague convention to interpret what we mean,
    – Stephen C
    13 hours ago
















  • 9




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    yesterday










  • @StephenC big-O notation implies "worst case".
    – David M
    17 hours ago










  • It doesn't always. For example, we don't say that a HashMap insertion is O(N)! By convention, big O refers to the "average" or "typical" case if you don't qualify it. The problem is that for the OP's example, it is difficult to say what the average is. That is why it is a good idea to be explicit, rather than relying on some vague convention to interpret what we mean,
    – Stephen C
    13 hours ago










9




9




In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
yesterday




In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
yesterday












@StephenC big-O notation implies "worst case".
– David M
17 hours ago




@StephenC big-O notation implies "worst case".
– David M
17 hours ago












It doesn't always. For example, we don't say that a HashMap insertion is O(N)! By convention, big O refers to the "average" or "typical" case if you don't qualify it. The problem is that for the OP's example, it is difficult to say what the average is. That is why it is a good idea to be explicit, rather than relying on some vague convention to interpret what we mean,
– Stephen C
13 hours ago






It doesn't always. For example, we don't say that a HashMap insertion is O(N)! By convention, big O refers to the "average" or "typical" case if you don't qualify it. The problem is that for the OP's example, it is difficult to say what the average is. That is why it is a good idea to be explicit, rather than relying on some vague convention to interpret what we mean,
– Stephen C
13 hours ago




















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